Question

A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point $A$? $(g=10m{s}^2$)
In the above problem, what is the component of its velocity perpendicular to the plane when it strikes at $A$?

Answer 1

For particle projected up the plane, in this situation we take two axes , one perpendicular to the plane and other along the plane then , taking component of velocity and acceleration along these axes we get :

along x-axis $u_x=10\cos {30}^0=5\sqrt{3},a_x=-g\sin {30}^0=-5$

and along y-axis, $u_y=10\sin {30}^0=5,a_y=-g\cos {30}^0=-5\sqrt{3}$

using these we get, time of flight $T=\frac{2u_y}{a_y}=\frac{2\times 5}{5\sqrt{3}}=\frac{2}{\sqrt{3}}$

so, $v_x=u_x+a_{x}t=5\sqrt{3}-5\times \frac{2}{\sqrt{3}}=\frac{5\sqrt{3}}{3}m/s$

and $v_y=u_y+a_{y}t=5-5\sqrt{3}\times \frac{2}{\sqrt{3}}=-5m/s$

so speed with which particle strike the inclined will be $v=\sqrt{(}{v}_x^2+v_x^2)=\sqrt{(}25/3+25)=\frac{10}{\sqrt{3}}m/s$

and component of velocity perpendicular to the plane when it strikes at A is $5m/s$